How do you solve this?

Question

1. 6 over n minus 1 over 2 equals 9 over 2n

2. 1/2 plus 2x minus 5 over 2x equals 14/x minus 1/3

3. 5/2 minus 3z -6 over 3z equals z - 8 over 2z

4. 10 -z over 4z equals 1 over z -1

5. 5c to the second power + 4 - 7 OVER 5c +8 equals
1/15(15c - 13)



Thank you for your time

Answer 1

(1)
6/n - 1/2 = 9/2n

put all the n's together then solve

6/n - 9/2n = 1/2

12/2n - 9/2n = 1/2

3/2n = 1/2

2n = 6

(2)
1/2 + 2x - 5/2x = 14/x - 1/3

hmmm, you need to use brackets in here

is this
1/2 + (2x-5)/2x = 14/x - 1/3

or 1/2 + (2x) - 5/2x = 14/x - 1/3

and there are other combinations too ...

1/2 + (2x-5)/2x = 14/x - 1/3
x/2x + (2x+5)/2x = 42/3x - x/3x
(3x+5)/2x = (42-x)/3x
this is gonna get messy

9x^2 + 15x = 84x - 2x^2
7x^2 - 69x = 0
x(7x - 69) = 0

x = 0 or 69/7

but is it really the question you asked?

(3)
5/2 - (3z-6)/3z = (z-8)/2z

you need brackets here too

you know ... just repost it with all the brackets so we know exactly what sums we are answering

looking back you could be asking a totally different thing for (1) even

(6/(n-1))/2 = 9/2n for instance - exceedingly messy, but possibly the sum you have described.

Answer 2

1.
6/n-1/2=9/2n
6/n-9/2n=1/2
(6*2)/2n-9/2n=1/2
12/2n-9/2n=1/2
3/2n=1/2
6=2n
n=3


2.
1/2+2x-5/2x=14/x-1/3
1/2+1/3+2x=14/x+5/2x
(1/2+1/3)+2x=(14*2)/2x+5/2x
(5/6)+2x=28/2x+5/2x
5/6+2x=33/2x
(5/6)(2x)+(2x*2x)=33
5x/3+4x^2=33
5x+12x^2=99
12x^2+5x-99=0
using formula:{[-b+/-(b^2-4ac)]/2a}
x=2.67 or -3.09(up to 3 significant figure)

3.
5/2-(3z-6)/3z=(z-8)/2z
(5*3z)/(2*3z)-[2(3z-6)/(3z*2)]=3(z-8)/(2z*3)
15z/6z-[(6z-12)/6z]=(3z-24)/6z
[15z-6z+12]/6z=(3z-24)/6z
(9z+12)/6z=(3z-24)6z
9z+12=3z-24
6z=-24-12
6z=-36
z=-6


4.
(10-z)/4z=1/(z-1)
(10-z)(z-1)=4z
10z-10-z^2+z=4z
11z-10-z^2=4z
z^2-11z+4z+10=0
z^2-7z+10=0
(z-2)(z-5)=0
z=2 or z=5

5.
I don't understand your question